原文

给定一个长度为n的整型数组，找出所有出现超过 ⌊ n/3 ⌋ 次的元素。算法应该运行在线性时间上，且进用O(1)空间。

提示：

它可能有多少个主要元素？

原文

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space.

Hint:

How many majority elements could it possibly have?

分析

The Boyer-Moore Vote Algorithm solves the majority vote problem in linear time O(n)and logarithmic space O(logn). The majority vote problem is to determine in any given sequence of choices whether there is a choice with more occurrences than half of the total number of choices in the sequence and if so, to determine this choice. Note how this definition contrasts with the mode in which it is not simply the choice with the most occurrences, but the number of occurrences relative to the total length of the sequence.

Mathematically, given a finite sequence (length n) of numbers, the object is to find the majority number defined as the number that appears more than ⌊n/2⌋ times.

import java.util.*;
public class MajorityVote {
     public int majorityElement(int[] num) {
         int n = num.length;
         int candidate = num[0], counter = 0;
         for (int i : num) {
             if (counter == 0) {
                 candidate = i;
                 counter = 1;
             } else if (candidate == i) {
                 counter++;
             } else {
                 counter--;
             }
        }

        counter = 0;
        for (int i : num) {
            if (i == candidate) counter++;
        }
        if (counter <= n / 2) return -1;
        return candidate;
    }
    public static void main(String[] args) {
        MajorityVote s = new MajorityVote();
        System.out.format("%d\n", s.majorityElement(new int[] {1, 2, 3}));
        System.out.format("%d\n", s.majorityElement(new int[] {2, 2, 3}));
    }
}

代码

Java

public class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> marjority = new ArrayList<>();
        int n = nums.length;
        int candidate1 = 0, candidate2 = 0, counter1 = 0, counter2 = 0;
        for (int i : nums) {
            if (candidate1 == i) {
                counter1++;
            } else if (candidate2 == i) {
                counter2++;
            } else if (counter1 == 0) {
                candidate1 = i;
                counter1 = 1;
            } else if (counter2 == 0) {
                candidate2 = i;
                counter2 = 1;
            } else {
                counter1--;
                counter2--;
            }
        }
        counter1 = 0;
        counter2 = 0;
        for (int i : nums) {
            if (i == candidate1) counter1++;
            else if (i == candidate2) counter2++;
        }
        if (counter1 > n/3) marjority.add(candidate1);
        if (counter2 > n/3) marjority.add(candidate2);
        return marjority;
    }
}